C++: Giving unknown class variable to template -

i've got class abc , want give unknown class variable template, that:

template < v > class abc { // }; 

of course above code doesn't work (there wasn't type of parameter v). have got ideas fix that? don't want give type of variable v template.

there 2 functions, types different. how use both functions in class template below?

#include <iostream> using namespace std; // types of functions max1 , min1 different! int max1(int a, int b){ return a>b?a:b; } int& min1(int a, int b){ return a<b?a:b; }  template<typename _t, _t(function)(_t,_t)> class abc {       public:       _t a, b;       _t get()       {                  return function(a, b);              }         };  abc <int, max1> abc; // <- if write "abc <int, min1> abc;", error comes! how fix this?  int main()     {     abc.a = 3;     abc.b = 8;     cout << abc.get() << '\n';           cin.get();           return 0;           } 

_{seems irrelevant, question unclear @ beginning...}

template < v > 

must be

template < typename v > 

or

template < class v > 

typename preferable.

---

regarding edit: problem is, min1 returns &. if remove it, may have:

#include <iostream>  int max1(int a, int b){ return a>b?a:b; } int min1(int a, int b){ return a<b?a:b; }  template<typename _t, _t(*function)(_t,_t)> class abc { public:       _t a, b;       _t get() { return function(a, b); }         };  int main() {     abc <int, max1> abc;     abc <int, min1 > def;      abc.a = def.a = 3;     abc.b = def.b = 8;      std::cout << abc.get() << '\n';     std::cout << def.get() << '\n';            return 0;       } 

and print 8 3